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Difference between revisions of "2017 AIME II Problems/Problem 3"

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Revision as of 01:39, 7 November 2020

Problem

A triangle has vertices $A(0,0)$, $B(12,0)$, and $C(8,10)$. The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution 1

[asy] pair A,B,C,D,X,Z,P; A=(0,0); B=(12,0); C=(8,10); X=(10,5); Z=(6,0); P=(6,3.4); fill(B--X--P--Z--cycle,lightgray); draw(A--B--C--cycle); dot(A); label("$A$",A,SW); dot(B); label("$B$",B,SE); dot(C); label("$C$",C,N); draw(X--P,dashed); draw(Z--P,dashed); dot(X); label("$X$",X,NE); dot(Z); label("$Z$",Z,S); dot(P); label("$P$",P,NW); [/asy]

The set of all points closer to point $B$ than to point $A$ lie to the right of the perpendicular bisector of $AB$ (line $PZ$ in the diagram), and the set of all points closer to point $B$ than to point $C$ lie below the perpendicular bisector of $BC$ (line $PX$ in the diagram). Therefore, the set of points inside the triangle that are closer to $B$ than to either vertex $A$ or vertex $C$ is bounded by quadrilateral $BXPZ$. Because $X$ is the midpoint of $BC$ and $Z$ is the midpoint of $AB$, $X=(10,5)$ and $Z=(6,0)$. The coordinates of point $P$ is the solution to the system of equations defined by lines $PX$ and $PZ$. Using the point-slope form of a linear equation and the fact that the slope of the line perpendicular to a line with slope $m$ is $-\frac{1}{m}$, the equation for line $PX$ is $y=\frac{2}{5}x+1$ and the equation for line $PZ$ is $x=6$. The solution of this system is $P=\left(6,\frac{17}{5}\right)$. Using the shoelace formula on quadrilateral $BXPZ$ and triangle $ABC$, the area of quadrilateral $BXPZ$ is $\frac{109}{5}$ and the area of triangle $ABC$ is $60$. Finally, the probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to vertex $A$ or vertex $C$ is the ratio of the area of quadrilateral $BXPZ$ to the area of $ABC$, which is $\frac{\frac{109}{5}}{60}=\frac{109}{300}$. The answer is $109+300=\boxed{409}$.

Solution 2

Since we know the coordinates of all three vertices of the triangle, we can find the side lengths: $AB=12$, $AC=2\sqrt{41}$, and $BC=2\sqrt{29}$. We notice that the point where the three distances are the same is the circumcenter - so we use one of the triangle area formulas to find the circumradius, since we know what the area is. \[\frac{12 \cdot 2\sqrt{41} \cdot 2\sqrt{29}}{4 \cdot R}=\frac{12 \cdot 10}{2}.\] We rearrange to get \[R=\frac{\sqrt{41} \cdot \sqrt{29}}{5}.\] [asy] draw((0,0)--(12,0)--(8,10)--(0,0)); draw((6,0)--(6,3.4)--(10,5)); draw((6,3.4)--(4,5)); label("$A$", (0,0), SW); label("$B$", (12,0), SE); label("$C$", (8, 10), N); label("$P$", (6, 3.4), NNE); label("$R$", (10, 5), NE); label("$S$", (6, 0), S); label("$T$", (4, 5), NW); [/asy] We know that $AP=\frac{\sqrt{41} \cdot \sqrt{29}}{5}$, and $AS=6$, so using the Pythagorean Theorem gives $SP=\frac{17}{5}$. This means $[ASP]=[BSP]=\frac{17}{5} \cdot 6 \cdot \frac{1}{2} = \frac{51}{5}$. Similarly, we know that $BP=\frac{\sqrt{41} \cdot \sqrt{29}}{5}$, and $BR=\sqrt{29}$, so we get that $PR=\frac{4\sqrt{29}}{5}$, and so $[BRP]=[CRP]=\frac{4\sqrt{29}}{5} \cdot \sqrt{29} \cdot \frac{1}{2} = \frac{58}{5}$. Lastly, we know that $CP=\frac{\sqrt{41} \cdot \sqrt{29}}{5}$, and $CT=\sqrt{41}$, so we get that $PT=\frac{2\sqrt{41}}{5}$, and $[ATP]=[CTP]=\frac{2\sqrt{41}}{5} \cdot \sqrt{41} \cdot \frac{1}{2} = \frac{41}{5}$. Therefore, our answer is $\frac{51+58}{2(51+58+41)}=\frac{109}{300} \rightarrow \boxed{409}$.

Solution 3

To start the problem, identify the two midpoints that connect $AB$ and $BC$. This is because the midpoints of such lines is the mark at which the point will sway closer to vertex $A$/$C$ or vertex $B$. The midpoint of $AB$ is $(6,0)$, and the midpoint of $BC$ is $(10,5)$. Then, determine the line at which the distance between vertex $B$ and vertex $C$ are the same. Assuming that $x$ is the real value of $x$ and $y$ is the real value of $y$, we can create a simple equation:

$\sqrt{|x-8|^2 + (10-y)^2}$ = $\sqrt{(12-x)^2 + y^2}$

where the left side of the equation is for the distance to vertex $C$ and the right side of the equation is the distance to vertex $B$.

Squaring both sides and then distributing, we get

$x^2 - 16x + 64 + y^2 - 20y + 100 = x^2 - 24x + 144 + y^2$.

Notice that $(x-8)^2 = (-x+8)^2$, and thus there is no need to create another equation.

Simplifying, we get $8x + 20 = 20y$.

Divide both sides by 20, then simplify, and the line that represents equivalent distance between vertex $B$ and vertex $C$ is $y = \frac{2x}{5} + 1$.

This line starts at the midpoint of $BC$, which is $(10,5)$, and ends at the line $x=6$, as $x=6$ represents equivalent distance between vertex $A$ and vertex $B$. Plug in $x=6$ to the equation $y = \frac{2x}{5} + 1$, and we get $y = \frac{17}{5}$. Now that we have our four points that are $(6,0), (6,\frac{17}{5}), (10,5)$, and $(12,0)$, we can calculate the area of the quadrilateral in which a point is closer to vertex $B$ as opposed to either vertex $A$ or vertex $C$. Simply draw a rectangle that has the points $(6,0), (6,5), (12,5)$ and $(12,0)$, and then subtract the two triangles that appear in between.

Thus, the area of the quadrilateral is $6\cdot5 - (\frac{\frac{8}{5}\cdot4}{2} + \frac{5\cdot2}{2}) \rightarrow 30 - \frac{41}{5} = \frac{109}{5}$. Since the problem asks us for the probability that a point chosen inside the triangle is inside the quadrilateral, and because the area of $\triangle ABC$ is $\frac{12\cdot10}{2} = 60$, the probability is $\frac{\frac{109}{5}}{60}=\frac{109}{300}$, which means the final answer is $109+300=\boxed{409}$.

Solution by IronicNinja~

Solution 4

See Also

2017 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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